Learning Log
Why
are you advised to open
windows slightly if a tornado
approaches?
Ch. 10 & 11 - Gases
Dalton’s Law
Ideal Gas Law
(p. 322-325, 340-346)
A. Dalton’s Law of Partial Pressures
 Total
pressure of a
mixture of gases is equal
to the sum of the partial
pressures of the
component gases.
• In the absence of a
chemical reaction
PT = P1+ P2+ P3+…
Practice Problem
A mixture of four gases exerts a total
pressure of 1200 mm Hg. Gases A and
B each exert 420 mm Hg. Gas C exerts
280 mm Hg. What pressure is exerted
by gas D?
PT = P1 + P2 + P3 + P4
1200 = 420 mm Hg + 420 mm Hg
+ 280 mm Hg + P4
1200 = 1120 mm Hg + P4
P4 = 80 mm Hg
B. Vapor pressure of water
 Gases
are often collected in lab by
water displacement and are mixed
with water vapor
 Patm = Pgas + PH2O
 To determine the pressure of the
gas collected – subtract the vapor
pressure of the water at that
temperature from the current
atmospheric pressure
C. Ideal Gas Law
 The
mathematical relationship
among pressure, volume,
temperature and the number of
moles of a gas.
 Derived by combining the gas laws.
PV=nRT
D. Ideal Gas Constant
Merge the Combined Gas Law with Avogadro’s Principle:
PV
V
k
=R
nT
T
n
IDEAL GAS CONSTANT
R=0.0821 Latm/molK
R=8.315 dm3kPa/molK
E. Ideal Gas Law Problems
 Calculate
the pressure in atmospheres
of 0.412 mol of He at 16°C & occupying
3.25 L.
GIVEN:
WORK:
P = ? atm
PV = nRT
n = 0.412 mol
P(3.25)=(0.412)(0.0821)(289)
L
mol Latm/molK K
T = 16°C = 289 K
V = 3.25 L
P = 3.01 atm
R = 0.0821Latm/molK
E. Ideal Gas Law Problems
 Find
the volume of 85 g of O2 at 25°C
and 104.5 kPa.
GIVEN:
WORK:
V=?
85 g 1 mol = 2.7 mol
n = 85 g = 2.7 mol
32.00 g
T = 25°C = 298 K PV = nRT
P = 104.5 kPa
(104.5)V=(2.7) (8.315) (298)
kPa
mol
dm3kPa/molK K
R = 8.315 dm3kPa/molK
V = 64 dm3
F. Finding Molar Mass
from the Ideal Gas Law
n =
mass
n =m
Molar mass
M
PV=nRT
PV = mRT
M
M = mRT
PV
OR
G. Finding Density
from the Ideal Gas Law
D = mass
volume
or
M = mRT
PV
M = DRT
P
D = MP
RT
D=m
V
PRACTICE PROBLEMS
 At
28°C and 0.974 at, 1.00 L of gas has
a mass of 5.16 g. What is the molar
mass of this gas?
P
= 0.974 atm
 T = 28°C = 273 = 301 K
M
V = 1.00 L
m = 5.16 g
= mRT
PV
= (5.16 g) (0.0821 L∙atm/mol∙K) (301K)
(0.974 atm)(1.00 L)
= 131g/mol
PRACTICE PROBLEMS
 What
is the density of a sample of
ammonia gas, NH3, if the pressure is
0.928 atm and the temperature is
63.0°C?

P = 0.928 atm
M = 17.034 g/mol

D = MP
RT
=
T = 63.0°C + 273 = 336 K
R = 0.0821 L∙atm/mol∙K
(17.034 g/mol)(0.928 atm)
(0.0821L∙atm/mol∙K)((336 K)
= 0.572 g/L NH3
Homework Assignment
Workbook.
Complete
problems #1 and 2
on pp. 173-174, 1-2 p. 178,
1-2 p. 180
Practice Test Part 2
 P.
181 – 182 #1-7, 12, 18 - 21
Practice test Part 2
 Workbook
p. 182 #1-7